Question-51733

Question Number 51733 by Tinkutara last updated on 30/Dec/18

Commented by maxmathsup by imad last updated on 30/Dec/18

let I =∫_0 ^(π/2) (dx/(a^2 cos^2 x +b^2 sin^2 x)) ⇒I =∫_0 ^(π/2) (dx/(a^2 ((1+cos(2x))/2) +b^2 ((1−cos(2x))/2))) =2 ∫_0 ^(π/2) (dx/(a^2 +b^2 +(a^2 −b^2 )cos(2x))) =_(2x=u) 2 ∫_0 ^π (du/(2( a^2 +b^2 +(a^2 −b^2 )cosu))) = ∫_0 ^π (du/(a^2 +b^2 +(a^2 −b^2 )cosu)) =_(tan((u/2))=t) ∫_0 ^∞ (1/(a^2 +b^2 +(a^2 −b^2 )((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^∞ ((2dt)/((a^2 +b^2 )(1+t^2 ) +(a^2 −b^2 )(1−t^2 ))) =∫_0 ^∞ ((2dt)/(2a^2 +2b^2 t^2 )) =∫_0 ^∞ (dt/(a^2 +b^2 t^2 )) =(1/a^2 )∫_0 ^∞ (dt/(1+((b/a)t)^2 ))) with ab>0 =_((b/a)t =α) (1/a^2 ) ∫_0 ^∞ (1/(1+α^2 )) (a/b) dα =(1/(ab)) ∫_0 ^∞ (dα/(1+α^2 )) =(1/(ab)).(π/2) =(π/(2ab)) . let f(a) = ∫_0 ^(π/2) (dx/(a^2 cos^2 x +b^2 sin^2 x)) ⇒f^′ (a) =∫_0 ^(π/2) ((−2acos^2 x)/((a^2 cos^2 x+b^2 sin^2 x)^2 )) =((−2)/a) ∫_0 ^(π/2) ((a^2 cos^2 x +b^2 sin^2 x−b^2 sin^2 x)/((a^2 cos^2 x +b^2 sin^2 x)^2 ))dx =−(2/a) ∫_0 ^π (dx/(a^2 cos^2 x+b^2 sin^2 x)) +((2b^2 )/a) ∫_0 ^(π/2) ((1−cos^2 x)/((a^2 cos^2 x +b^2 sin^2 x)^2 ))dx =−(2/a)f(a) +((2b^2 )/a) ∫_0 ^(π/2) (dx/((a^2 cos^2 x +b^2 sin^2 x)^2 )) +(b^2 /a^2 ) ∫_0 ^(π/2) ((−2acos^2 x)/((a^2 cos^2 x +b^2 sin^2 x)^2 ))dx f^′ (a) =−(2/a)f(a) +((2b^2 )/a) ∫_0 ^(π/2) (...)dx+(b^2 /a^2 )f^′ (a) ⇒ (1−(b^2 /a^2 ))f^′ (a) +(2/a)f(a) =((2b^2 )/a) ∫_0 ^(π/2) (dx/((a^2 cos^2 x +b^2 sin^2 x)^2 )) ⇒ (((a^2 −b^2 )f^′ (a)+2af(a))/a^2 ) =((2b^2 )/a) ∫_0 ^(π/2) (dx/((a^2 cos^2 x +b^2 sin^2 x)^2 )) ⇒ ∫_0 ^(π/2) (dx/((a^2 cos^2 x +b^2 sin^2 x)^2 )) =(a/(2b^2 )) (1/a^2 ){(a^2 −b^2 )f^′ (a)+2a f(a)} = (1/(2ab^2 )){(a^2 −b^2 )((π/(2b))(((−1)/a^2 )) +2a (π/(2ab))} =(1/(2ab^2 )){ −(π/(2a^2 b))(a^2 −b^2 ) +(π/b)} a=2 and b=3 ⇒ ∫_0 ^(π/2) (dx/((4cos^2 x +9sin^2 x)^2 )) =(1/(4.9)){((π(5))/(8.3)) +(π/3)} =(1/(36)){((5π)/(24)) +((8π)/(24))} =(1/(36)) .((13π)/(24)) =((13π)/(36.24)) =((13π)/(864))

l e t I = \int_{0}^{\frac{π}{2}} \frac{d x}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x} \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{d x}{a^{2} \frac{1 + c o s (2 x)}{2} + b^{2} \frac{1 - c o s (2 x)}{2}}

= 2 \int_{0}^{\frac{π}{2}} \frac{d x}{a^{2} + b^{2} + (a^{2} - b^{2}) c o s (2 x)} =_{2 x = u} 2 \int_{0}^{π} \frac{d u}{2 (a^{2} + b^{2} + (a^{2} - b^{2}) c o s u)}

= \int_{0}^{π} \frac{d u}{a^{2} + b^{2} + (a^{2} - b^{2}) c o s u} =_{t a n (\frac{u}{2}) = t} \int_{0}^{\infty} \frac{1}{a^{2} + b^{2} + (a^{2} - b^{2}) \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

= \int_{0}^{\infty} \frac{2 d t}{(a^{2} + b^{2}) (1 + t^{2}) + (a^{2} - b^{2}) (1 - t^{2})}

= \int_{0}^{\infty} \frac{2 d t}{2 a^{2} + 2 b^{2} t^{2}} = \int_{0}^{\infty} \frac{d t}{a^{2} + b^{2} t^{2}} = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{d t}{1 + {(\frac{b}{a} t)}^{2})} w i t h a b > 0

=_{\frac{b}{a} t = α} \frac{1}{a^{2}} \int_{0}^{\infty} \frac{1}{1 + α^{2}} \frac{a}{b} d α = \frac{1}{a b} \int_{0}^{\infty} \frac{d α}{1 + α^{2}} = \frac{1}{a b} . \frac{π}{2} = \frac{π}{2 a b} .

l e t f (a) = \int_{0}^{\frac{π}{2}} \frac{d x}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x} \Rightarrow f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{- 2 {a c o s}^{2} x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}}

= \frac{- 2}{a} \int_{0}^{\frac{π}{2}} \frac{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x - b^{2} {s i n}^{2} x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} d x

= - \frac{2}{a} \int_{0}^{π} \frac{d x}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x} + \frac{2 b^{2}}{a} \int_{0}^{\frac{π}{2}} \frac{1 - {c o s}^{2} x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} d x

= - \frac{2}{a} f (a) + \frac{2 b^{2}}{a} \int_{0}^{\frac{π}{2}} \frac{d x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} + \frac{b^{2}}{a^{2}} \int_{0}^{\frac{π}{2}} \frac{- 2 {a c o s}^{2} x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} d x

f^{^{'}} (a) = - \frac{2}{a} f (a) + \frac{2 b^{2}}{a} \int_{0}^{\frac{π}{2}} (\dots) d x + \frac{b^{2}}{a^{2}} f^{^{'}} (a) \Rightarrow

(1 - \frac{b^{2}}{a^{2}}) f^{^{'}} (a) + \frac{2}{a} f (a) = \frac{2 b^{2}}{a} \int_{0}^{\frac{π}{2}} \frac{d x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} \Rightarrow

\frac{(a^{2} - b^{2}) f^{^{'}} (a) + 2 a f (a)}{a^{2}} = \frac{2 b^{2}}{a} \int_{0}^{\frac{π}{2}} \frac{d x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{d x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} = \frac{a}{2 b^{2}} \frac{1}{a^{2}} {(a^{2} - b^{2}) f^{^{'}} (a) + 2 a f (a)}

= \frac{1}{2 {a b}^{2}} {(a^{2} - b^{2}) (\frac{π}{2 b} (\frac{- 1}{a^{2}}) + 2 a \frac{π}{2 a b}}

= \frac{1}{2 {a b}^{2}} {- \frac{π}{2 a^{2} b} (a^{2} - b^{2}) + \frac{π}{b}}

a = 2 a n d b = 3 \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{d x}{{(4 {c o s}^{2} x + 9 {s i n}^{2} x)}^{2}} = \frac{1}{4.9} {\frac{π (5)}{8.3} + \frac{π}{3}} = \frac{1}{36} {\frac{5 π}{24} + \frac{8 π}{24}}

= \frac{1}{36} . \frac{13 π}{24} = \frac{13 π}{36.24} = \frac{13 π}{864}

Commented by Tinkutara last updated on 30/Dec/18

Thanks Sir!

Commented by turbo msup by abdo last updated on 30/Dec/18

y o u a r e w e l c o m e s i r .

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

∫_0 ^(π/2) (dx/(a^2 cos^2 x+b^2 sin^2 x)) ∫_0 ^(π/2) ((sec^2 xdx)/(a^2 +b^2 tan^2 x)) (1/b^2 )∫_0 ^(π/2) ((d(tanx))/(((a/b))^2 +tan^2 x)) (1/b^2 )×(1/(((a/b))))∣tan^(−1) (((tanx)/(a/b)))∣_0 ^(π/2) (1/(ab))[tan^(−1) (((tan(π/2))/(a/b)))−tan^(−1) ((0/(a/b)))] =(1/(ab))×tan^(−1) (∞)=(1/(ab))×(π/2)=(π/(2ab)) solving method−1 ∫_0 ^(π/2) (dx/((a^2 cos^2 x+b^2 sin^2 x)))=(π/(2ab)) diferentiate both side w.r.t a ∫_0 ^(π/2) (∂/∂a)[(/((a^2 cos^2 x+b^2 sin^2 x)))]dx=(π/(2b))×((−1)/a^2 ) ∫_0 ^(π/2) ((−1×2acos^2 x)/((a^2 cos^2 x+b^2 sin^2 x)^2 ))dx=((−π)/(2a^2 b)) ∫_0 ^(π/2) ((cos^2 xdx)/((a^2 cos^2 x+b^2 sin^2 x)^2 ))dx=(π/(4a^3 b)).....eqn p again diffdrentiate w.r.t b ∫_0 ^(π/2) ((−2bsin^2 x)/((a^2 cos^2 x+b^2 sin^2 x)^2 ))dx=((−π)/(2ab^2 )) ∫_0 ^(π/2) ((sin^2 x)/((a^2 cos^2 x+b^2 sin^2 x)^2 ))=(π/(4ab^3 )).....eqn q add eqn p and q ∫_0 ^(π/2) ((cos^2 x+sin^2 x)/((a^2 cos^2 x+b^2 sin^2 x)^2 ))dx=(π/(4ab))((1/a^2 )+(1/b^2 ))=((π(a^2 +b^2 ))/(4a^3 b^3 )) so answer for given problem a=2 b=3 =((π(4+9))/(4×8×27))=((13π)/(864))

\int_{0}^{\frac{π}{2}} \frac{d x}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x}

\int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} x d x}{a^{2} + b^{2} {t a n}^{2} x}

\frac{1}{b^{2}} \int_{0}^{\frac{π}{2}} \frac{d (t a n x)}{{(\frac{a}{b})}^{2} + {t a n}^{2} x}

\frac{1}{b^{2}} \times \frac{1}{(\frac{a}{b})} ∣ {t a n}^{- 1} (\frac{t a n x}{\frac{a}{b}}) ∣_{0}^{\frac{π}{2}}

\frac{1}{a b} [{t a n}^{- 1} (\frac{t a n \frac{π}{2}}{\frac{a}{b}}) - {t a n}^{- 1} (\frac{0}{\frac{a}{b}})]

= \frac{1}{a b} \times {t a n}^{- 1} (\infty) = \frac{1}{a b} \times \frac{π}{2} = \frac{π}{2 a b}

s o l v i n g m e t h o d - 1

\int_{0}^{\frac{π}{2}} \frac{d x}{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)} = \frac{π}{2 a b}

d i f e r e n t i a t e b o t h s i d e w . r . t a

\int_{0}^{\frac{π}{2}} \frac{\partial}{\partial a} [\frac{}{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}] d x = \frac{π}{2 b} \times \frac{- 1}{a^{2}}

\int_{0}^{\frac{π}{2}} \frac{- 1 \times 2 {a c o s}^{2} x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} d x = \frac{- π}{2 a^{2} b}

\int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} x d x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} d x = \frac{π}{4 a^{3} b} \dots . . e q n p

a g a i n d i f f d r e n t i a t e w . r . t b

\int_{0}^{\frac{π}{2}} \frac{- 2 {b s i n}^{2} x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} d x = \frac{- π}{2 {a b}^{2}}

\int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} = \frac{π}{4 {a b}^{3}} \dots . . e q n q

a d d e q n p a n d q

\int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} x + {s i n}^{2} x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} d x = \frac{π}{4 a b} (\frac{1}{a^{2}} + \frac{1}{b^{2}}) = \frac{π (a^{2} + b^{2})}{4 a^{3} b^{3}}

s o a n s w e r

f o r g i v e n p r o b l e m a = 2 b = 3

= \frac{π (4 + 9)}{4 \times 8 \times 27} = \frac{13 π}{864}

Commented by mr W last updated on 30/Dec/18

sir, the question asks ∫_0 ^(π/2) (dx/((a^2 cos^2 x+b^2 sin^2 x)^2 ))=?

s i r, t h e q u e s t i o n a s k s

\int_{0}^{\frac{π}{2}} \frac{d x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}} = ?

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

o k \dots

l e t m e s o l v e \dots

Commented by mr W last updated on 30/Dec/18

y e s, y o u c a n!

Commented by mr W last updated on 30/Dec/18

v e r y n i c e m e t h o d s i r!

Commented by Tinkutara last updated on 30/Dec/18

Can we differentiate like this w.r.t any variable?

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

y e s \dots

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

ok i am solving the problem in another method and upload problems with solution similar method of method 1

o k i a m s o l v i n g t h e p r o b l e m i n a n o t h e r m e t h o d

a n d u p l o a d p r o b l e m s w i t h s o l u t i o n s i m i l a r m e t h o d

o f m e t h o d 1

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

t h a n k y o u s i r \dots

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

t h a n k y o u f o r e x c e l l e n t p o s t \dots

Commented by Tinkutara last updated on 30/Dec/18

Thank you very much Sir! I got the answer. ��

Commented by afachri last updated on 30/Dec/18

i like your solution above Mr Tanmay. i save it to observe the steps. amazing

i like your solution above Mr Tanmay .

i save it to observe the steps . amazing

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

method 2 ∫_0 ^(π/2) (dx/((a^2 cos^2 x+b^2 sin^2 x)^2 )) ∫_0 ^(π/2) ((sec^4 xdx)/(b^4 [((a/b))^2 +tan^2 x]^2 )) now put tanx=(a/b)tank sec^2 xdx=(a/b)sec^2 kdk ∫_0 ^(π/2) (((1+(a^2 /b^2 )tan^2 k)×(a/b)sec^2 kdk)/(b^4 [((a/b))^2 +((a/b))^2 tan^2 k]^2 )) (a/b^3 )×(1/b^4 )∫_0 ^(π/2) (((b^2 +a^2 tan^2 k)sec^2 kdk)/(((a/b))^4 sec^4 k)) (a/b^7 )×(b^4 /a^4 )∫_0 ^(π/2) ((b^2 /(sec^2 k))+((a^2 tan^2 k)/(sec^2 k)))dk =(1/(a^3 b^3 ))∫_0 ^(π/2) (b^2 cos^2 k+a^2 sin^2 k)dk (1/(a^3 b^3 ))×(1/2)∫_0 ^(π/2) b^2 (1+cos2k)+a^2 (1−cos2k)dk (1/(2a^3 b^3 ))[∣b^2 (k+((sin2k)/2))+a^2 (k−((sin2k)/2))∣_0 ^(π/2) ] (1/(2a^3 b^3 ))[b^2 ((π/2)+((sinπ)/2))+a^2 ((π/2)−((sinπ)/2))] =(1/(2a^3 b^3 ))[(π/2)(a^2 +b^2 )] [sinπ=0] =((π(a^2 +b^2 ))/(4a^3 b^3 ))=((π(2^2 +3^2 ))/(4×8×27))=((13π)/(864))

m e t h o d 2

\int_{0}^{\frac{π}{2}} \frac{d x}{{(a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x)}^{2}}

\int_{0}^{\frac{π}{2}} \frac{{s e c}^{4} x d x}{b^{4} {[{(\frac{a}{b})}^{2} + {t a n}^{2} x]}^{2}}

n o w p u t t a n x = \frac{a}{b} t a n k

{s e c}^{2} x d x = \frac{a}{b} {s e c}^{2} k d k

\int_{0}^{\frac{π}{2}} \frac{(1 + \frac{a^{2}}{b^{2}} {t a n}^{2} k) \times \frac{a}{b} {s e c}^{2} k d k}{b^{4} {[{(\frac{a}{b})}^{2} + {(\frac{a}{b})}^{2} {t a n}^{2} k]}^{2}}

\frac{a}{b^{3}} \times \frac{1}{b^{4}} \int_{0}^{\frac{π}{2}} \frac{(b^{2} + a^{2} {t a n}^{2} k) {s e c}^{2} k d k}{{(\frac{a}{b})}^{4} {s e c}^{4} k}

\frac{a}{b^{7}} \times \frac{b^{4}}{a^{4}} \int_{0}^{\frac{π}{2}} (\frac{b^{2}}{{s e c}^{2} k} + \frac{a^{2} {t a n}^{2} k}{{s e c}^{2} k}) d k

= \frac{1}{a^{3} b^{3}} \int_{0}^{\frac{π}{2}} (b^{2} {c o s}^{2} k + a^{2} {s i n}^{2} k) d k

\frac{1}{a^{3} b^{3}} \times \frac{1}{2} \int_{0}^{\frac{π}{2}} b^{2} (1 + c o s 2 k) + a^{2} (1 - c o s 2 k) d k

\frac{1}{2 a^{3} b^{3}} [∣ b^{2} (k + \frac{s i n 2 k}{2}) + a^{2} (k - \frac{s i n 2 k}{2}) ∣_{0}^{\frac{π}{2}}]

\frac{1}{2 a^{3} b^{3}} [b^{2} (\frac{π}{2} + \frac{s i n π}{2}) + a^{2} (\frac{π}{2} - \frac{s i n π}{2})]

= \frac{1}{2 a^{3} b^{3}} [\frac{π}{2} (a^{2} + b^{2})] [s i n π = 0]

= \frac{π (a^{2} + b^{2})}{4 a^{3} b^{3}} = \frac{π (2^{2} + 3^{2})}{4 \times 8 \times 27} = \frac{13 π}{864}

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